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 one year ago
The x coordinate of a point on the curve 9y^2 = x^3 , the normal at which cuts off equal intercept on the axes is ?
 one year ago
The x coordinate of a point on the curve 9y^2 = x^3 , the normal at which cuts off equal intercept on the axes is ?

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sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.19y^2=x^3 d(9y^2)/dx =d(x^3)/dx 18y dy/dx =3x^2 dy/dx =x^2/(6y)

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.1The gradient of tangent is x^2/(6y) So, gradient of normal is 6y/(x^2)

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.1NOW, let xintercept of the normal be a then yintercept is also a.

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.1Gradient of normal = (a0)/(0a) 6y/(x^2) =1

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.1Now,dw:1359201825424:dw and solve for x

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.1dw:1359201971611:dw

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.1U need to solve that

Yahoo!
 one year ago
Best ResponseYou've already chosen the best response.06y/(x^2) =1......then wat did u do ?

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.1Substitute y interms of x
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